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  • Can someone explain - Electromagnetic Induction and Alternating currents - JEE Main-2

In a series LCR circuit R = 200 \Omega and the voltage and the frequency of the main supply is 220 V and 50 Hz respectively. On taking out the capacitance from the circuit the current lags behind the voltage by 300. On taking out the inductor from the circuit the current leads the voltage by 300. The power dissipated in the LCR circuit is

  • Option 1)

    242 W

  • Option 2)

    305 W

  • Option 3)

    210 W

  • Option 4)

    zero W

 

Answers (1)

best_answer

As we learnt in 

R.L Circuit Voltage -

V_{R}= IR

V_{L}= {I}X_{L}

- wherein

 

 

RC Circuit Voltage -

V_{R}=IR

V_{C}={I}X_{C}

- wherein

 

 R=500\Omega

\phi =30^{\cdot }(When capacitance is taken out)

tan30^{^{\circ}}=\frac{\mathrm{X_L} }{\mathrm{R} }---------------(1)

on taking out inductor, current lead by 30^{\circ}

\therefore tan 30^{\circ}=\frac{\mathrm{Xc} }{\mathrm{R} }-----------------------(2)

X_{L}=X_{C}=\frac{\mathrm{R} }{\mathrm{\sqrt{3}} }

\therefore z=R

power dissipated in the circuit=\frac{\mathrm{V^{2}} }{\mathrm{R}}=\frac{\mathrm{220^{2}} }{\mathrm{200}}=242 W


Option 1)

242 W

This option is correct

Option 2)

305 W

This option is incorrect

Option 3)

210 W

This option is incorrect

Option 4)

zero W

This option is incorrect

Posted by

divya.saini

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